AMC 10 · 2014 · #21

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
pythagorean-theoremsystems-of-equationsarea-triangles convert-to-algebra ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 4 insights
Problem
Trapezoid ABCD has parallel sides AB = 33 and CD = 21, and its two slanted legs measure 10 and 14. Angles A and B (at the long base) are acute. Find the length of the shorter of the two diagonals AC and BD.

Pick an answer.

(A)
$10\sqrt{6}$
(B)
25
(C)
$8\sqrt{10}$
(D)
$18\sqrt{2}$
(E)
26

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

A trapezoid with two given bases and two given legs is begging for coordinates, so Tool #1 (Draw a Diagram) is primary: lay the long base AB on the x-axis and drop perpendiculars from the top corners. Tool #4 (Introduce a Variable) names the two horizontal offsets p and q and the height h. Tool #7 (Identify Subproblems) turns each leg into its own right triangle via the Pythagorean theorem. Tool #13 (Convert to Algebra) then solves the little system for p, q, h, after which the two diagonals are just distances between corner points.

1STEP 1

Put the trapezoid on a grid

Lay the long base flat: A = (0, 0), B = (33, 0), with D = (p, h) and C = (p + 21, h) on top, so DC is parallel and 21 long.

A=(0,0), B=(33,0), D=(p, h), C=(p+21, h)
2STEP 2

Name the two base offsets

The bases differ by 33 - 21 = 12, split between left overhang p and right overhang q; acute angles keep both positive, so p + q = 12.

p + q = 33 - 21 = 12
3STEP 3

Pythagoras on each leg

Each leg is a hypotenuse over run p or q and rise h, with 10 on the left and 14 on the right: p² + h² = 100 and q² + h² = 196.

p² + h² = 100, q² + h² = 196
4STEP 4

Solve for the offsets and height

Subtracting kills h²: (q - p)(12) = 96 so q - p = 8, and with q + p = 12 that gives q = 10, p = 2, h² = 96.

q² - p² = 96 → (q-p)(12)=96 → q-p=8; q=10, p=2, h²=96
5STEP 5

Measure both diagonals

So D = (2, h), C = (23, h): AC² = 23² + 96 = 625 gives AC = 25, while BD² = 31² + 96 = 1057, about 32.5 — so AC is shorter.

AC² = 23² + 96 = 625 → AC = 25; BD² = 31² + 96 = 1057 → BD ≈ 32.5
Answer
25
The height works out to h = √(96) = 4√6 ≈ 9.8, comfortably less than both legs 10 and 14, which is required since a leg is always at least as long as the height it spans — good. The diagonal AC = 25 reaches from a bottom corner across to the far top corner; 25 is longer than the longer leg 14 and shorter than the bottom base 33, exactly the range you expect for a diagonal. BD ≈ 32.5 is clearly the longer one because D sits almost directly above A, leaving BD to span nearly the whole base. So the shorter diagonal is 25, matching (B). The distractors 10√6 ≈ 24.5, 8√(10) ≈ 25.3, 18√2 ≈ 25.5, and 26 all cluster near 25 to punish arithmetic slips, but only the exact 625 = 25² lands on a clean value.
💡Key takeaway

Stand the trapezoid on the x-axis, split each slanted leg into a right triangle, and the extra 33 - 21 = 12 of base splits as 2 and 10; then diagonal AC = √(23² + 96) = 25 is the short one.

  • Put the trapezoid on a grid
  • Name the two base offsets
  • Pythagoras on each leg
  • Solve for the offsets and height
  • Measure both diagonals

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